Leetcode 117: Populating Next Right Pointers in Each Node II-优快云博客

本文介绍了一种在任意二叉树中填充每个节点的Next指针的方法，确保相同层级的相邻节点通过Next指针连接。该解决方案仅使用常数额外空间，并遵循从右到左的处理顺序。

Follow up for problem "Populating Next Right Pointers in Each Node".

What if the given tree could be any binary tree? Would your previous solution still work?

Note:

You may only use constant extra space.

For example,
Given the following binary tree,

After calling your function, the tree should look like:

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \    \
    4-> 5 -> 7 -> NULL


Note: must process right tree first.

 1 /**
 2  * Definition for binary tree with next pointer.
 3  * struct TreeLinkNode {
 4  *  int val;
 5  *  TreeLinkNode *left, *right, *next;
 6  *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     void connect(TreeLinkNode *root) {
12         if (root == NULL) return;        
13         
14         if (root->left != NULL && root->right != NULL) root->left->next = root->right;
15         
16         TreeLinkNode* cur = root->right != NULL ? root->right : root->left;
17         
18         if (cur != NULL)
19         {
20             TreeLinkNode* next = root->next;
21             
22             while (next != NULL)
23             {
24                 if (next->left != NULL)
25                 {
26                     cur->next = next->left;
27                     break;
28                 }
29                 
30                 if (next->right != NULL)
31                 {
32                     cur->next = next->right;
33                     break;
34                 }
35                 
36                 next = next->next;
37             }
38         }
39         
40         connect(root->right);
41         connect(root->left);
42     }
43 };